View Full Version : physics problem relating to Energy...i feel dumb


biggy35
11-14-06, 01:03 PM
http://aycu05.webshots.com/image/5524/2001559919200289681_rs.jpg (http://allyoucanupload.webshots.com/v/2001559919200289681)

For this first question...I cannot firgure out something simple:

Using conservation of energy, find the speed Vb of the block at the bottom of the ramp. (Express your answer in terms of some or all the variables m , v , h and and any appropriate constants.)

so this is asking for an expression and not an equation. I know that it includes v, h, g. I have tried sqrte(gh/v) and many other things but i cannot firgure it out and my professor doesnt teach! Somone please help me!

and problem #2..(i feel dumb again)

http://aycu38.webshots.com/image/7357/2001502235031555526_rs.jpg (http://allyoucanupload.webshots.com/v/2001502235031555526)

What is the initial kinetic energy of the skier? What is the final kinetic energy of the skier? (Enter expressions for the initial and final kinetic energy separated by a comma in terms of the variables given in the problem.)

And i dont know the right way to put this...someone please help again! thanks

James R
11-14-06, 06:01 PM
For the first problem, I'll assume the surfaces are frictionless.

The total energy E of the block at any point is:

E = U + K

where U is the gravitational potential energy and K is the kinetic energy.

We also have:

U = mgh

where m is the mass, g is the gravitational acceleration and h is the height.

And

K = (1/2)mv^2

is the kinetic energy.

When the block is at the top of the ramp its total energy is therefore

E = mgH + (1/2)mv^2

where H is the height of the ramp and v is its speed at the top.

At the bottom of the ramp, the energy is:

E = 0 + (1/2)m v_f^2

where v_f is the final speed, and there is no potential energy.

Energy is conserved in this example, since the system of Earth+block+ramps is isolated, so the E's are constant in both cases. Therefore:

mgH + (1/2)m v^2 = (1/2) m v_f^2

Re-arranging to find v_f, we get:

2gH + v^2 = v_f^2

or

v_f = sqrt(v^2 + 2gH)

Does that help?

biggy35
11-14-06, 06:57 PM
Thanks so much! that makes so much more sense now. Anyone have any thought to offer for the skier?

James R
11-14-06, 06:59 PM
Well, kinetic energy is K=(1/2)mv^2, so...

Billy T
11-14-06, 08:09 PM
Just looked at his pictures and block on ramp can be made into interesting problem:

In terms of angles, ramp height, with no friction and square block of uniform density, what are the conditions when it hits floor without tiping over as he has it shown. Conversely what conditions just make it tip?

It is late and I must go to bed, but quickly think (first impression) this may be an "insoluable problem" if floor is rigid with infinte forces coming into play.

biggy35
11-14-06, 09:25 PM
So since kinetic energy is that formula. I have still tried different attempts with: 1/2mv_i^2, 1/2mv_f^2

But i cant firgure it out. And i do want everyone to know that this is my first physics class ever haha. So im sorry im so dumb

James R
11-14-06, 11:00 PM
I don't know if you've posted the entire question for problem 2. All I see is a diagram.

CANGAS
11-15-06, 12:37 AM
So since kinetic energy is that formula. I have still tried different attempts with: 1/2mv_i^2, 1/2mv_f^2

But i cant firgure it out. And i do want everyone to know that this is my first physics class ever haha. So im sorry im so dumb


haha haha haha bein dum is easy ansern yur qwestin is wat got them stump

beddy by

CANGAS
11-15-06, 12:47 AM
http://aycu05.webshots.com/image/5524/2001559919200289681_rs.jpg (http://allyoucanupload.webshots.com/v/2001559919200289681)

For this first question...I cannot firgure out something simple:

Using conservation of energy, find the speed Vb of the block at the bottom of the ramp. (Express your answer in terms of some or all the variables m , v , h and and any appropriate constants.)

so this is asking for an expression and not an equation. I know that it includes v, h, g. I have tried sqrte(gh/v) and many other things but i cannot firgure it out and my professor doesnt teach! Somone please help me!

and problem #2..(i feel dumb again)

http://aycu38.webshots.com/image/7357/2001502235031555526_rs.jpg (http://allyoucanupload.webshots.com/v/2001502235031555526)

What is the initial kinetic energy of the skier? What is the final kinetic energy of the skier? (Enter expressions for the initial and final kinetic energy separated by a comma in terms of the variables given in the problem.)

And i dont know the right way to put this...someone please help again! thanks

you want exact numbers an not eqwasons but yu don giv all informatin abot inital conditon?

FOO on yu!

biggy35
11-15-06, 02:24 AM
So the problem states that:

Marjan, a skier of mass 'm', coasts a distance 's' on level snow to a stop from a speed of 'v', as depicted in the figure. Use the work-energy principle to find the coefficient of kinetic friction 'uk' (upside down h) between the skis and the snow.

By looking at my skier diagram...

What is the initial kinetic energy of the skier? What is the final kinetic energy of the skier? Enter expressions for the initial and final kinetic energy separated by a comma in terms of the variables given in the problem.

So im pretty sure that i am not calculating a number at all. It asks directly for an expression of the initial kinetic energy and final kinetic energy with the terms listed above for you. Can anyone firgure it out now with all this information??

CANGAS
11-15-06, 02:53 AM
wot planet we ar on? wot gee ar we in? how far way from sinter of mas? etc? etc? etc? etc?

we gas kweston an anser tu?

hord enuf tu gas anser wen we areddy no kweston.

ho tu gas anser wen yu don tel kweston?

CANGAS
11-15-06, 02:59 AM
[QUOTE=
so this is asking for an expression and not an equation. [/QUOTE]

Is it correct to assume, according to the usual use of the language, that when you demand a "expression", you are demanding a specific number for an answer?

biggy35
11-15-06, 03:59 AM
haha i dont know. im confused....but i have problems reading some of the stuff you write

biggy35
11-15-06, 10:04 AM
and it says answer this by using an expression with the variables...so i would assume this is not going to be like just like two numbers.

§outh§tar
11-15-06, 01:27 PM
biggy35, why dont you try solving for the coefficient of friction using newtons laws and kinematics instead and then tell us what you get.